解析

//! n,log(n)~n->每个节点访问1次，共n次;函数递归栈占用log(n)，极端不平衡树的函数递归栈占用n
class Solution {
 public:
  bool isSymmetric(TreeNode *root) {
    TreeNode *node = root;
    return dfsJudge(node->left, node->right);
  }
  bool dfsJudge(TreeNode *nodeLeft, TreeNode *nodeRight) {
    if (nodeLeft == nullptr && nodeRight == nullptr) {
      return true;
    } else if (nodeLeft == nullptr || nodeRight == nullptr) {
      return false;
    }
    bool a = dfsJudge(nodeLeft->left, nodeRight->right);
    bool b = dfsJudge(nodeLeft->right, nodeRight->left);
    return a && b && (nodeLeft->val == nodeRight->val);
  }
};